Question

The radius of a 10 inch right circular cylinder is measured to be 4 inches, but with a possible error of ±0.1 inch. What is the resulting possible error in the volume of the cylinder? Include units in your answer

Answer 1

Answer:

Possible error in the volume of the cylinder is ±8π inch³.

Step-by-step explanation:

We have, a right circular cylinder with radius, r = 4 inches and height, h = 10 inches.

As, the volume of a cylinder = $\pi r^{2} h$

So, volume of the given cylinder, V = $10\pi r^{2}$

Then, $\frac{dV}{dr}=20\pi r$ i.e. $dV=(20\pi r)dr$,

where dV and dr represents the possible change (or error) in volume and radius of the cylinder respectively.

As, we have,

r = 4 inches and dr = ±0.1 inch.

Substituting the values, we get,

$dV=(20\pi r)dr$

⇒  $dV=20\pi 4\times \pm 0.1$

⇒  $dV=80\pi \times \pm 0.1$

⇒  $dV=\pm 8\pi$ inch³

Hence, the possible error in the volume of the cylinder is ±8π inch³.